let [root(n 2 +1)]=[root(n 2 +λ)] , where n, λ€N . show that λ can have 2n different values.

Question

let [root(n 2 +1)]=[root(n 2 +&lambda;)] , where n, &lambda;&euro;N . show that &lambda; can have 2n different values.

Arun Kumar · Answer

Hello

Well for [root(n^2+1)] =n for any n greater than 0

n^2+lamda should between n^2 and (n+1)^2

=> (n+1)^2-n^2=2n+1 if we include 0

So it can have 2n+1 different values

take 2,3

9-4=5 which is 2*2+1 if we include 0

Thanks & Regards
Arun Kumar
Btech IIT delhi
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let [root(n 2 +1)]=[root(n 2 +λ)] , where n, λ€N . show that λ can have 2n different values.

let [root(n²+1)]=[root(n²+λ)] , where n, λ€N . show that λ can have 2n different values.

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let [root(n 2 +1)]=[root(n 2 +λ)] , where n, λ€N . show that λ can have 2n different values.

let [root(n2+1)]=[root(n2+λ)] , where n, λ€N . show that λ can have 2n different values.

1 Answers

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Other Related Questions on Differential Calculus

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let [root(n²+1)]=[root(n²+λ)] , where n, λ€N . show that λ can have 2n different values.